A nice cuberoot simplification| Math Olympiad
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- čas přidán 9. 09. 2024
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The method is very much available in the text books. Then why in CZcams? To find a shortcut. We have 26 +15r(3). First of all make sure that the quantity in the radical is a prime. That is 3 here in this case. Put y = 3 and 3x^2+y = 15 --> x = + or - 2. As -2+r(3) is negative, answer is 2+ r(3)
Try extracting qube root from 37+30*sqrt(3) using your method. You will fail.
If you change your method a little and look for the answer in the form a+b*sqrt(3) - you will succeed.
Numbers that are not square free are the cause of your problem.
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There is a good way to simplify the task. If you want to take the kth root of
a+b*sqrt(c), first calculate the norm, n=a^2-c*(b^2). If n = x^k with an integer x, you have a chance, but no guarantee of success, to take the root and its norm will be x. If n!=x^k - you have no chance of getting an answer in the form of quadratic irrationality.
@@ald6980 brother. There is no substitute for traditional method. If there were a substitute for traditional method, the genius will write the same in the text books and we would have learnt the same. For example, division by 10; division by 5; division by 3; etc. Conclusion is NO SUBSTITUTE FOR THE TRADITIONAL METHOD. Now the question is why you tube? when something is available in text books? To find a silly shortcut. A shortcut is not a formula for every problem of that type. There are many ifs in a shortcut. If a shortcut is TRUE for every problem then it becomes a theorem. OK?
Now cr(37+30r(3)) is as per short cut 1+r(3) which is wrong. Now include one more small prime in root. That is r(3x2). Hence the problem reduces to
cr(37 +15r(6)) for which the answer is y=6 and 3x^2 +y = 15 --> x =1. So final answer is cr(37+30r3)) = 1+r(6). Note that here 37 is a prime and hence the difficulty in shortcut
Проще будет если а+с*sqrt 3
Возводим в куб
а^3+3с^3*sqrt 3+3a^2*c*sqrt 3+3*3c^2*a
Легко видно что
а*(а^2+9с^2)=26
с(9с^2+3а^2)=15